\newproblem{lay:2_1_22}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.1.22}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Show that if the columns of $B$ are linearly independent, so are the columns of $AB$.
}{
  % Solution
	This statement is not true. For instance, the columns of
	\begin{center}
		$B=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}$
	\end{center}
	are linearly independent. However, given $A=\begin{pmatrix}1 & 2\\2 & 4\end{pmatrix}$, the columns of
	\begin{center}
		$AB=\begin{pmatrix}1 & 2\\2 & 4\end{pmatrix}\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}=\begin{pmatrix}1 & 2\\2 & 4\end{pmatrix}$
	\end{center}
	are not linearly independent because the second column is twice the first one.
	
	$AB$ is linearly independent if the columns of $A$ and $B$ are linearly independent.
}
\useproblem{lay:2_1_22}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
